Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 555;
int g[MAXN][MAXN];
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  scanf("%d%d", &n, &m);
  for (int(i) = 0; (i) < (int)(n); i++)
    for (int(j) = 0; (j) < (int)(m); j++) scanf("%d", &g[i][j]);
  int best = -1;
  vector<int> best_a, best_b;
  for (int(x) = 0; (x) < (int)(MAXN); x++) {
    vector<int> a(n), b(m);
    a[0] = x;
    for (int(j) = 0; (j) < (int)(m); j++) b[j] = g[0][j] - x;
    for (int(i) = 0; (i) < (int)(n); i++) a[i] = g[i][0] - b[0];
    bool ok = true;
    for (int(i) = 0; (i) < (int)(n); i++) ok &= (a[i] >= 0);
    for (int(j) = 0; (j) < (int)(m); j++) ok &= (b[j] >= 0);
    for (int(i) = 0; (i) < (int)(n); i++)
      for (int(j) = 0; (j) < (int)(m); j++) ok &= (a[i] + b[j] == g[i][j]);
    if (ok) {
      int total =
          accumulate(a.begin(), a.end(), 0) + accumulate(b.begin(), b.end(), 0);
      if (best == -1 || total < best) {
        best = total;
        best_a = a;
        best_b = b;
      }
    }
  }
  cout << best << endl;
  if (best != -1) {
    for (int(i) = 0; (i) < (int)(n); i++)
      for (int(j) = 0; (j) < (int)(best_a[i]); j++) printf("row %d\n", i + 1);
    for (int(j) = 0; (j) < (int)(m); j++)
      for (int(i) = 0; (i) < (int)(best_b[j]); i++) printf("col %d\n", j + 1);
  }
  return 0;
}